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# IEEE floating point representation

This presentation was made for student batch 2017-2018 of MBSTU. Here we will get
IEEE 32 bit floating representation .
IEEE 754 floating point representation

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### IEEE floating point representation

1. 1. Computer Organization And Architecture Presented by :Maskur Al Shal Sabil ID: IT18021 Dept : Information & Communication Technology Mawlana Bhashani Science & Technology University 10/20/2020 1IT18021
2. 2. Learning Outcome • Floating Point Representation • IEEE 754 Standards For Floating Point Representation • Single Precision • Double Precision • Single Precision Addition 10/20/2020 IT18021 2
3. 3. Floating Point Representation The floating point representation does not reserve any specific number of bits for the integer part or the fractional part. Instead it reserve a certain point for the number and a certain number of bit where within that number the decimal place sits called the exponent. 10/20/2020 IT18021 3
4. 4. IEEE 754 Floating point representation According to IEEE754 standard, the floating point number is represented in following ways: • Half Precision(16bit):1 sign bit,5 bit exponent & 10 bit mantissa • Single Precision(32bit):1 sign bit,8 bit exponent & 23 bit mantissa • Double Precision(64bit):1 sign bit,11 bit exponent & 52bit mantissa • Extend precision(128bit):1 sign bit,15bit exponent & 112 bit mantissa 10/20/2020 IT18021 4
5. 5. Floating Point Representation 10/20/2020 IT18021 5 The floating point representation has two part : the one signed part called the mantissa and other called the exponent. (sign) × mantissa × 2exponent Sign Bit Exponent Mantissa
6. 6. Decimal To Binary Conversion 10/20/2020 IT18021 6 (55.35)10 = (?)2 (55)10=(110111)2 (0.35)10 = (010110)2 (45.45)10=(110111.010110)2 32 16 8 4 2 1 1 1 0 1 1 1 0.35 × 2 0 .7 0.7× 2 1 .4 .4 × 2 0 .8 .8× 2 1 .6 .6 × 2 1 .2 .2× 2 0 .4
7. 7. Scientific Notation - 1.602 ×10-19 sign significand Base Exponent 10/20/2020 IT18021 7
8. 8. IEEE 32-bit floating point representation 10/20/2020 IT18021 8 1-bit 8 -bit 23- bit Number representation: (-1)S × 1.M× 2E-127 Sign Bit Biased Exponent Trailing Significand bit or Mantissa
9. 9. IEEE 32-bit floating point representation (45.45)10=(101101.011100)2 Step -1: Normalize the number Step-2: Take the exponent and mantissa. Step-3:Find. the bias exponent by adding 127 Step-3:Normalize the mantissa by adding 1. Step -4:Set the sign bit 0 if positive otherwise 1 . For n bit exponent bias is 2n-1-1 10/20/2020 IT18021 9
10. 10. IEEE 32-bit floating point representation 10/20/2020 IT18021 10 (45.45)10 = (?)2 (45)10=(101101)2 (0.45)10 = (011100)2 (45.45)10=(101101.011100)2 32 16 8 4 2 1 1 0 1 1 0 1 0.45 × 2 0 .9 0.9 × 2 1 .8 .8 × 2 1 .6 .6 × 2 1 .2 .2 × 2 0 .4 .4 × 2 0 .8
11. 11. IEEE 32-bit floating point representation (45.45)10=(101101.011100)2 101101.011100 = 1.01101011100 × 25 Here bias exponent = 5 + 127 = 132 mantissa=01101011100 1-bit 8 -bit 23- bit 10/20/2020 IT18021 11 Sign Bit Biased Exponent Trailling Significand bit or Mantissa
12. 12. IEEE 32-bit floating point representation (132)10=(?)2 128 64 32 16 8 4 2 1 1 0 0 0 0 1 0 0 (132)10=(10000100)2 10/20/2020 IT18021 12 0 10000100 01101011100110011001100 1-bit 8 -bit 23- bit
13. 13. IEEE 64-bit floating point representation 1bit 11bits 52bits Here we use 211-1 – 1 = 1023 as bias value. 10/20/2020 IT18021 13 Sign Bit Biased Exponent Trailling Significand bit or Mantissa
14. 14. IEEE 64-bit floating point representation (45.45)10=(101101.011100)2 101101.011100 = 1.01101011100 × 25 Here bias exponent = 5 + 1023=1028= (10000000100)2 mantissa=01101011100 1-bit 11 -bits 52- bits 10/20/2020 IT18021 14 0 10000000100 01101011100110011001100……
15. 15. Convert Floating Point To Decimal 0100 0000 0100 0110 1011 0000 0000 0000 exponent Mantissa Number representation: (-1)S × 1.M× 2E-127 S=0 E=(1000000)2=(64) 10 M =(.100 0110 1011 0000 0000 0000 )2= (0.5537109375)10 (-1)0 × 1.5537109375 × 2 64-127 = 1.68453677×10−19 10/20/2020 IT18021 15
16. 16. Addition of floating point First consider addition in base 10 if exponent is the same the just add the significand 5.0E+2 +7.0E+2 12.0E+2=1.2E+3 10/20/2020 IT18021 16
17. 17. Addition of floating point 1.2232E+3 + 4.211E+5 First Normalize to higher exponent a. Find the difference between exponents b. Shift smaller number right by that amount 1.2232E+3=.012232E+5 10/20/2020 IT18021 17
18. 18. Addition of floating point 4.211 E+5 + 0.012232 E+5 4.223232 E+5 10/20/2020 IT18021 18
19. 19. 32Bit floating point addition a 0 1101 0111 111 0011 1010 0000 1100 0011 b 0 1101 0111 000 1110 0101 1111 0001 1100 Find the 32 bit floating point number representation of a+b . Here, e=(11010111)= (215)10 m= (111 0011 1010 0000 1100 0011) 10/20/2020 IT18021 19
20. 20. 32Bit floating point addition a= (-1)0 × 1. 111 0011 1010 0000 1100 0011 × 2127-215 =1.111 0011 1010 0000 1100 0011 × 212 e=(11010111)= (215)10 m= 000 1110 0101 1111 0001 1100 b= 1. 000 1110 0101 1111 0001 1100 × 212 + a= 1.111 0011 1010 0000 1100 0011 × 212 11 . 000 0 001 1111 1111 1101 1111 × 212 10/20/2020 IT18021 20